Question

One mole of iron (6 x 10^23 atoms) has a mass of 56 grams, and its density is 7.87 grams per cubic centimeter, so the center-to-center distance between atoms is 2.28 x 10^-10 m. You have a long thin bar of iron, 2.3 m long, with a square cross section, 0.12 cm on a side. You hang the rod vertically and attach a 149 kg mass to the bottom, and you observe that the bar becomes 1.17 cm longer. From these measurements, it is possible to determine the stiffness of one interatomic bond in iron.
ks = 20014.1 N/m
Number of side-by-side long chains of atoms = 4.81e^12
Number of bonds in total length = 1.096e^10

What is the stiffness of a single interatomic "spring"?

Answer 1

Answer:

Explanation:

Given that:

length l = 2.3 m

a = 0.12 cm = $0.12 \times 10^{-2} \ m$

$x = 1.17 \ cm = 1.17 \times 10^{-2}\ m$

m = 149 kg

$\delta = 7.87 \ g/cm^3$

$da = 2.28 \times 10^{-10}\ m$

$F_{net} = F-mg\\ \\0 = F - mg \\ \\ F = mg \\ \\ k_sx = mg$

∴

$k_s = \dfrac{149(9.8)}{1.17 \times 10^{-2}} \\ \\ k_s = 124803.42 \ N /m$

$N_{chain} = \dfrac{A_{wire}}{A_{atom}} = \dfrac{A_w}{da^2}$

$N_{chain} = \dfrac{(a)^2}{(da)^2} = (\dfrac{a}{da})^2$

$N_{chain} = (\dfrac{0.12 \times 10^{-2} }{2.28 \times 10^{-10}})^2$

$N_{chain} = 2.77 \times 10^{13}$

$N_{bond} = \dfrac{L}{da} \\ \\ = \dfrac{2.3}{2.28 \times 10^{-10}} \\ \\ N_{bond} = 1.009 \times 10^{10}$

$\text{Finally; the stiffness of a single interatomic spring is:}$

$k_{si} =\dfrac{N_{bond}}{N_{chain}}\times k_s$

$k_{si} =\dfrac{(1.009 \times 10^{10})}{2.77*10^{13}}}\times (124803.42)$

$\mathbf{k_{si} =45.46 \ N/m}$